1. ## Integrate cos(x)cos(4x) dx

I'm trying to integrate:

$\displaystyle \int\cos(x)\cos(4x) dx$

Could some one tell me what I should be seeing about this to help me solve it. I thought I needed to change the $\displaystyle 4x$ into $\displaystyle x$ and tried doing that using the double angle formula, but I have to use it twice and end up with a integrand that looks too complicated. I'd like to know what a good first step would be.

Thank you.

2. ## Re: Integrate cos(x)cos(4x) dx

this is easy...but you have to use a formula that transforms the product cosxcos4x into a sum.

HERE cos(4x)cos(x)=(1/2)[cos(5x)-cos(3x)}

Minoas

3. ## Re: Integrate cos(x)cos(4x) dx

Originally Posted by Furyan
I'm trying to integrate:

$\displaystyle \int\cos(x)\cos(4x) dx$

Could some one tell me what I should be seeing about this to help me solve it. I thought I needed to change the $\displaystyle 4x$ into $\displaystyle x$ and tried doing that using the double angle formula, but I have to use it twice and end up with a integrand that looks too complicated. I'd like to know what a good first step would be.

Thank you.
\displaystyle \displaystyle \begin{align*} I &= \int{\cos{(x)}\cos{(4x)}\,dx} \\ I &= \frac{1}{4}\cos{(x)}\sin{(4x)} - \int{ -\frac{1}{4}\sin{(x)}\sin{(4x)}\,dx } \\ I &= \frac{1}{4}\cos{(x)}\sin{(4x)} + \frac{1}{4}\int{\sin{(x)}\sin{(4x)}\,dx} \\ I &= \frac{1}{4}\cos{(x)}\sin{(4x)} + \frac{1}{4} \left[ -\frac{1}{4}\sin{(x)}\cos{(4x)} - \int{ -\frac{1}{4}\cos{(x)}\cos{(4x)} \,dx} \right] \\ I &= \frac{1}{4}\cos{(x)}\sin{(4x)} - \frac{1}{16}\sin{(x)}\cos{(4x)} + \frac{1}{16}I \\ \frac{15}{16}I &= \frac{1}{4}\cos{(x)}\sin{(4x)} - \frac{1}{16}\sin{(x)}\cos{(4x)} \\ I &= \frac{4}{15}\cos{(x)}\sin{(4x)} - \frac{1}{15}\sin{(x)}\cos{(4x)} + C \end{align*}

4. ## Re: Integrate cos(x)cos(4x) dx

Thank you Prove It,

I haven't seen that before, integrating by parts twice and then letting the second integrand be I and solving that way. That's a very useful way of solving difficult integration problems. Thank you very much for showing me that. I tried this method with a similar problem that I had solved another way and found that I had to be very careful with signs and factors. I kept getting the wrong answer because I was getting a sign and factor wrong during the second integration. So I'm going to have to work on it, but it's great to know that if I can't see what else to do I can try this method.

Thanks again for all your efforts

5. ## Re: Integrate cos(x)cos(4x) dx

Originally Posted by Furyan
I'm trying to integrate:

$\displaystyle \int\cos(x)\cos(4x) dx$

Could some one tell me what I should be seeing about this to help me solve it. I thought I needed to change the $\displaystyle 4x$ into $\displaystyle x$ and tried doing that using the double angle formula, but I have to use it twice and end up with a integrand that looks too complicated. I'd like to know what a good first step would be.

Thank you.
Hi Furyan!

As you can see here, we have the identity:
$\displaystyle \cos \theta \cos \varphi = {{\cos(\theta - \varphi) + \cos(\theta + \varphi)} \over 2}$

In other words:
$\displaystyle \int\cos(4x)\cos(x) dx = \int{{\cos(4x - x) + \cos(4x + x)} \over 2} = \int{{\cos(3x) + \cos(5x)} \over 2}$

This is what Minoas already suggested (except for the typo ).

6. ## Re: Integrate cos(x)cos(4x) dx

Hi ILikeSerena

Thank you very much for that. I did, in fact, also solve this problem using the factor formula, as Minoanman suggested, and it was very much simpler. However, although there are only four factor formulae in my book I'm worried that I might not remember them in an exam. Although I found Prove It's method more difficult I feel I'm more likely to remember it and at least get some method marks, even if I make some mistakes.

I'm amazed how different the solutions look, depending on which method you use and yet they're equivalent. I solved one question using Prove It's method, the factor formula and a simple identity. All the solutions were equivalent, but they looked very different. I wouldn't have been able to rewrite any of them in terms of any of the others.

7. ## Re: Integrate cos(x)cos(4x) dx

Here's a trick that I like to use when I can't properly remember the formulas.

The only formula you need to remember is Euler's formula:
$\displaystyle e^{ix} = \cos x + i \sin x$

together with its rewritten forms (that you can both deduce from Euler's formula if you forget):
$\displaystyle \cos x = \dfrac 1 2 (e^{ix} + e^{-ix})$
$\displaystyle \sin x = \dfrac 1 {2i} (e^{ix} - e^{-ix})$

$\displaystyle \cos(4x)\cos(x) = \frac 1 2 (e^{i4x} + e^{-i4x}) \cdot \frac 1 2 (e^{ix} + e^{-ix})$
$\displaystyle = \frac 1 4 ((e^{i5x} + e^{-i5x}) + (e^{i3x} + e^{-i3x}))$

$\displaystyle = \frac 1 2 (\cos(5x) + \cos(3x)) \qquad \blacksquare$

8. ## Re: Integrate cos(x)cos(4x) dx

whoa! $\displaystyle \blacksquare$

That's a whole new level, but I'm liking it. I'm much better at deducing formualae than I am at remember them. Euler's formula, eh! I'm sure I came across that earlier today, but I glazed over when I saw $\displaystyle e^{ix}$. Is that i ,the i, $\displaystyle \sqrt{-1}?$. I'm going to look that up and try and find out why it's so fundamental. It looks very interesting.

Thank you very much for that

9. ## Re: Integrate cos(x)cos(4x) dx

Originally Posted by Furyan
whoa!

That's a whole new level, but I'm liking it. I'm much better at deducing formualae than I am at remember them. Euler's formula, eh! I'm sure I came across that earlier today, but I glazed over when I saw $\displaystyle e^{ix}$. Is that i ,the i, $\displaystyle \sqrt{-1}?$. I'm going to look that up and try and find out why it's so fundamental. It looks very interesting.

Thank you very much for that
Yep. It is that $\displaystyle i$.

According to wiki:
Euler's formula is ubiquitous in mathematics, physics, and engineering. The physicist Richard Feynman called the equation "our jewel" and "one of the most remarkable, almost astounding, formulas in all of mathematics."[2]

10. ## Re: Integrate cos(x)cos(4x) dx

Originally Posted by ILikeSerena
Yep. It is that $\displaystyle i$.

According to wiki:
Euler's formula is ubiquitous in mathematics, physics, and engineering. The physicist Richard Feynman called the equation "our jewel" and "one of the most remarkable, almost astounding, formulas in all of mathematics."[2]
Wow, it looks like a jewel. How exciting.

Thank you very much

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